双调和函数梯度范数的一个估计

An Estimation of the Gradient Norm of Biharmonic Functions

摘要: 双调和函数在数学界具有重要地位，且在现实中有广泛的应用。本文主要探究的是双调和函数梯度范数的一个估计，通过分析双调和函数与双解析函数的关系，计算出双调和函数的Poisson核，由此给出有界双调和函数梯度范数的一个估计，得到的积分表达式为今后进一步探究双调和函数的Khavinson猜想打下基础。

Abstract: Biharmonic functions play an important role in the field of mathematics and have wide applications in reality. This article mainly explores an estimate of the gradient norm of biharmonic functions. By analyzing the relationship between biharmonic functions and bianalytic functions, the Poisson kernel of biharmonic functions is calculated, and an estimate of the gradient norm of bounded biharmonic functions is given. The obtained integral expression lays the foundation for further exploration of the Khavinson conjecture of biharmonic functions in the future.

1. 基础知识

设D为单位圆盘。下面双调和方程 [1]

${\Delta }^2\varnothing \left(r,\theta \right)=\Delta \Delta \varnothing \left(r,\theta \right)=\frac{{\partial }^4\varnothing \left(r,\theta \right)}{\partial x^4}+2\frac{{\partial }^4\varnothing \left(r,\theta \right)}{\partial x^2\partial y^2}+\frac{{\partial }^4\varnothing \left(r,\theta \right)}{\partial y^4}=0$ (1)

$\forall 0\le r\le 1,0\le \theta <2\pi$

0在极坐标下，对于单位圆盘内的双调和函数：

$\varnothing \left(r,\theta \right)=\left(r^2-1\right)U_1\left(r,\theta \right)+U_2\left(r,\theta \right)$ (2)

这里的 $U_1\left(r,\theta \right)$ 和 $U_2\left(r,\theta \right)$ 分别都是调和函数。

我们首先将Dirichlet边界条件

${\varnothing \left(r,\theta \right)|}_{r=1}=\mu \left(\theta \right)$ (3)

代入上式，得到

${\varnothing \left(r,\theta \right)|}_{r=1}={\left(1^2-1\right)U_1\left(r,\theta \right)+U_2\left(r,\theta \right)|}_{r=1}={U_2\left(\theta \right)|}_{r=1}=\mu \left(\theta \right)$ (4)

上式表明关于调和函数 $U_2\left(r,\theta \right)$ 的Dirichlet边界值 [2] [3] 已求得，由调和函数的Poisson公式得到

$U_2\left(r,\theta \right)=\frac{1}{2\pi }{\displaystyle {\int }_0^{2\pi }\frac{\left(1-r^2\right)\mu \left(\psi \right)}{1-2r\cos \left(\theta -\psi \right)+r^2}\text{d}\psi }$ (5)

另一方面，两边关于r分别求偏导，然后再将Neumann边界条件

${\frac{\partial \varnothing }{\partial r}|}_{r=1}=0$ (6)

代入之，则有

${\frac{\partial \varnothing \left(r,\theta \right)}{\partial r}|}_{r=1}={\left[2U_1\left(r,\theta \right)+\frac{\partial U_2\left(r,\theta \right)}{\partial r}\right]|}_{r=1}=0$ (7)

又因为 $U_2\left(r,\theta \right)$ 对r求导得

$\begin{array}{c}{\frac{\partial U_2\left(r,\theta \right)}{\partial r}|}_{r=1}=\frac{1}{2\pi }{\displaystyle {\int }_0^{2\pi }\frac{-2\mu \left(\psi \right)\left[-2\cos \left(\theta -\psi \right)+2r\right]}{{\left[-2\cos \left(\theta -\psi \right)+2r\right]}^2}\text{d}\psi }\\ =-\frac{1}{2\pi }{\displaystyle {\int }_0^{2\pi }\frac{\mu \left(\psi \right)}{1-\cos \left(\theta -\psi \right)}\text{d}\psi }\end{array}$

将其代入，得到

${U_1\left(r,\theta \right)|}_{r=1}=\frac{1}{4\text{π}}{\displaystyle {\int }_0^{2\text{π}}\frac{\mu \left(\psi \right)}{1-\cos \left(\theta -\psi \right)}\text{d}\psi }$ (8)

再次对调和函数 $U_1\left(r,\theta \right)$ 用Poisson公式，得到

$\begin{array}{c}{U}_1\left(r,\theta \right)=\frac{1}{4\text{π}}{\displaystyle {\int }_0^{2\text{π}}\frac{\frac{1-r^2}{2\text{π}}{\displaystyle {\int }_0^{2\text{π}}\frac{\mu \left(\psi \right)}{1-\cos \left(\omega -\psi \right)}\text{d}\psi }}{1-2r\cos \left(\theta -\psi \right)+r^2}\text{d}\omega }\\ =\frac{1-r^2}{8{\text{π}}^2}{\displaystyle {\int }_0^{2\text{π}}\mu \left(\psi \right)\text{d}\psi }{\displaystyle {\int }_0^{2\text{π}}\frac{1}{\left(1-\cos \left(\omega -\psi \right)\right)\left[1-2r\cos \left(\theta -\omega \right)+r^2\right]}\text{d}\omega }\end{array}$

于是我们代入得到双调和函数满足边界条件的解为

$\begin{array}{l}U\left(z\right)=-\frac{{\left(r^2-1\right)}^2}{8{\text{π}}^2}{\displaystyle {\int }_0^{2\text{π}}\mu \left(\psi \right)\text{d}\psi }{\displaystyle {\int }_0^{2\text{π}}\frac{1}{\left[1-\cos \left(\omega -\psi \right)\right]\left[1-2r\cos \left(\theta -\omega \right)+r^2\right]}\text{d}\omega }\\ +\frac{1-r^2}{2\text{π}}{\displaystyle {\int }_0^{2\text{π}}\frac{\mu \left(\psi \right)}{1-2r\cos \left(\theta -\psi \right)+r^2}\text{d}\psi },\forall 0\le r\le 1,0\le \theta <2\text{π}\end{array}$ (9)

这里 $z=r{\text{e}}^{i\theta }$ ， $\zeta ={\text{e}}^{i\psi }$ 。

定义1令 $\zeta ={\text{e}}^{i\psi }$ ， $z=r{\text{e}}^{i\theta }$ ，双调和函数的Poisson核定义为

$P\left(z,\zeta \right)=-\frac{{\left(r^2-1\right)}^2}{4\text{π}}{\displaystyle {\int }_0^{2\text{π}}\frac{1}{\left[1-\cos \left(\omega -\psi \right)\right]\left[1-2r\cos \left(\theta -\omega \right)+r^2\right]}\text{d}\omega }+\frac{1-r^2}{{\left|z-\zeta \right|}^2},0\le r\le 1,0\le \theta <2\text{π}$ (10)

其中 $z=x+iy=r\cos \theta +ir\sin \theta$ ， $\left(x,y\right)\in D$ 。

设 $U\left(z\right)$ 是单位圆盘D上的有界双调和函数，则对几乎所有的 $\zeta \in \partial D$ ，都有径向边界值 [4] ：

$U^{*}\left(\zeta \right)={\lim }_{r\to 1^{-}}U\left(r\zeta \right)$ (11)

且 $U\left(z\right)$ 可以用 $U^{*}\left(\zeta \right)$ 的Poisson积分表示

$U\left(z\right)=P\left[U^{*}\right]\left(z\right)={\displaystyle {\int }_{\partial D}P\left(z,U^{*}\left(\zeta \right)\right)U^{*}\left(U^{*}\left(\zeta \right)\right)\text{d}s\left(U^{*}\left(\zeta \right)\right)}$ (12)

其中

$P\left(z,\zeta \right)=-\frac{{\left(r^2-1\right)}^2}{4\text{π}}{\displaystyle {\int }_0^{2\text{π}}\frac{1}{\left[1-\cos \left(\omega -\psi \right)\right]\left[1-2r\cos \left(\theta -\omega \right)+r^2\right]}\text{d}\omega }+\frac{1-r^2}{{\left|z-\zeta \right|}^2}$ (13)

对于 $z\in D$ ， $l\in \partial D$ ，由(12)有

$\langle \nabla U\left(z\right),l\rangle ={\displaystyle {\int }_{\partial D}\langle \nabla P\left(z,\zeta \right),l\rangle U^{*}\left(\zeta \right)\text{d}s\left(\zeta \right)}$ . (14)

设(14)左侧为 $\partial D$ 上本性有界函数空间的有界线性泛函Λ。由于Poisson算子是D上的本性有界双调和函数空间到 $\partial D$ 上本性有界函数 [5] 的空间的等距同构，我们也可以将Λ看成D上的本性有界双调和函数空间上的泛函，故成立

$\Vert \Lambda \Vert =C\left(z,l\right)$ , (15)

其中 $C\left(z,l\right)$ 为满足

$\left|\langle \nabla u,l\rangle \right|\le C\left(z,l\right){\sup }_{w\in D}\left|u\left(w\right)\right|$

的最佳系数 [6] 。因此，我们有

$C\left(z,l\right)={\displaystyle {\int }_{\partial D}\left|\langle \nabla P\left(z,\zeta \right),l\rangle \right|\text{d}s\left(\zeta \right)}$ (16)

2. C(z, l)的积分表达式

定理2.1

$\begin{array}{l}{\displaystyle {\int }_0^{2\pi }\frac{1}{\left[1-\cos \left(\omega -\psi \right)\right]\left[1-2r\cos \left(\theta -\omega \right)+r^2\right]}\text{d}\omega }\\ =\frac{2\pi r\cos \left(\psi -\theta \right)\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]-4\pi r^2{\sin }^2\left(\psi -\theta \right)}{{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^2\left(1-r^2\right)}\end{array}$ (17)

证明：根据文献 [7] 含参数奇异积分的导数性质 [7] ，我们有

$\begin{array}{c}\frac{\partial }{\partial \psi }{\displaystyle {\int }_0^{2\pi }\frac{\cot \frac{\omega -\psi }{2}}{1-2r\cos \left(\theta -\omega \right)+r^2}\text{d}\omega }={\displaystyle {\int }_0^{2\pi }\frac{\frac{\partial }{\partial \psi }\cot \frac{\omega -\psi }{2}}{1-2r\cos \left(\theta -\omega \right)+r^2}\text{d}\omega }\\ ={\displaystyle {\int }_0^{2\pi }\frac{1}{\left[1-\cos \left(\omega -\psi \right)\right]\left[1-2r\cos \left(\theta -\omega \right)+r^2\right]}\text{d}\omega }\end{array}$ (18)

令 $t=-\cot \frac{\omega -\psi }{2}$ ，得

$\frac{\cot \frac{\omega -\psi }{2}\text{d}\omega }{1-2r\cos \left(\theta -\omega \right)+r^2}=\frac{t\text{d}t}{1-2r\cos \left(\psi -\theta \right)+r^2-4rt\sin \left(\psi -\theta \right)+\left(1-2r\cos \left(\psi -\theta \right)+r^2\right)t^2}$

代入，得到

${\displaystyle {\int }_0^{2\pi }\frac{\cot \frac{\omega -\psi }{2}}{1-2r\cos \left(\theta -\omega \right)+r^2}\text{d}\omega }=\frac{2\pi r\sin \left(\psi -\theta \right)}{\left(1-2r\cos \left(\psi -\theta \right)+r^2\right)\left(1-r^2\right)}$ (19)

所以

$\begin{array}{l}\frac{\partial }{\partial \psi }{\displaystyle {\int }_0^{2\pi }\frac{\cot \frac{\omega -\psi }{2}}{1-2r\cos \left(\theta -\omega \right)+r^2}\text{d}\omega }\\ =\frac{2\pi r\cos \left(\psi -\theta \right)\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]-4\pi r^2{\sin }^2\left(\psi -\theta \right)}{{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^2\left(1-r^2\right)}\end{array}$

那么

$\begin{array}{c}P\left(r,\theta \right)=-\frac{1-r^2}{2}\left\{\frac{r\cos \left(\psi -\theta \right)\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]-2r^2{\sin }^2\left(\psi -\theta \right)}{{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^2}\right\}\\ +\frac{1-r^2}{1-2r\cos \left(\psi -\theta \right)+r^2}\end{array}$ (20)

定理2.2设 $\zeta ={\text{e}}^{i\psi }$ ， $x+iy=r{\text{e}}^{i\theta }$ ，当 $l=\left(1,0\right)$ 时，有

$\begin{array}{c}\langle \nabla P\left(z,\zeta \right),l\rangle =\frac{-3r^3{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2\cos \theta +r^5\cos \theta \left(1+r^2-{\left|z-\zeta \right|}^2\right)}{4{\left|z-\zeta \right|}^6}\\ -\frac{4r\cos \theta {\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}{4{\left|z-\zeta \right|}^6}+\frac{27r\cos \theta \left(1+r^2-{\left|z-\zeta \right|}^2\right)}{4{\left|z-\zeta \right|}^6}\\ +\frac{17r^3\cos \theta \left(1+r^2-{\left|z-\zeta \right|}^2\right)-5\left[\frac{{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}{r}\right]\cos \theta }{4{\left|z-\zeta \right|}^6}\\ +\frac{-40r^3\cos \theta +3\left(\frac{1+r^2-{\left|z-\zeta \right|}^2}{r}\right)\cos \theta -4r\cos \theta }{4{\left|z-\zeta \right|}^6}\end{array}$

$\begin{array}{l}+\frac{\frac{r^2-1}{r}\sin \theta \sqrt{4r^2-{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}\left[5\left(1+r^2-{\left|z-\zeta \right|}^2\right)\right]}{4{\left|z-\zeta \right|}^6}\\ +\frac{\frac{r^2-1}{r}\sin \theta \sqrt{4r^2-{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}\left[r^2\left(1+r^2-{\left|z-\zeta \right|}^2\right)\right]}{4{\left|z-\zeta \right|}^6}\\ +\frac{\frac{r^2-1}{r}\sin \theta \sqrt{4r^2-{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}\left(r^4-10r^2-3\right)}{4{\left|z-\zeta \right|}^6}\\ 0\le r\le 1,0\le \theta <2\pi \end{array}$ (21)

证明：双调和函数的Poisson核(20)的表达式为：

$\begin{array}{c}P\left(r,\theta \right)=-\frac{1-r^2}{2}\left\{\frac{r\cos \left(\psi -\theta \right)\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]-2r^2{\sin }^2\left(\psi -\theta \right)}{{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^2}\right\}\\ +\frac{1-r^2}{1-2r\cos \left(\psi -\theta \right)+r^2}\end{array}$

一、首先，分别对x和y对偏导数，表达式为

$\{\begin{array}{c}\frac{\partial P}{\partial x}=\frac{\partial P}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial P}{\partial \theta }\frac{\partial \theta }{\partial x}①\\ \frac{\partial P}{\partial y}=\frac{\partial P}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial P}{\partial \theta }\frac{\partial \theta }{\partial y}②\end{array}$

对 $\frac{\partial P}{\partial \theta }$ 和 $\frac{\partial P}{\partial r}$ 求解：

$\frac{\partial P}{\partial \theta }=\frac{\left(r^2-1\right)r\sin \left(\psi -\theta \right)\left[2r^3\cos \left(\psi -\theta \right)+r^4+10r\cos \left(\psi -\theta \right)-10r^2-3\right]}{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}$ (22)

和

$\begin{array}{c}\frac{\partial P}{\partial r}=\frac{27r^2\cos \left(\psi -\theta \right)-20r^3+3\cos \left(\psi -\theta \right)-4r}{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\\ +\frac{17r^4\cos \left(\psi -\theta \right)-10r{\cos }^2\left(\psi -\theta \right)}{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\\ +\frac{-6r^5{\cos }^2\left(\psi -\theta \right)+r^6\cos \left(\psi -\theta \right)-8r^3{\cos }^2\left(\psi -\theta \right)}{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\end{array}$ (23)

由链式法则，我们有

$\frac{\partial P}{\partial x}=\frac{\partial P}{\partial r}\cos \theta +\frac{\partial P}{\partial \theta }\left(-\frac{\sin \theta }{r}\right)=\frac{\partial P}{\partial r}\cos \theta -\frac{\partial P}{\partial \theta }\left(\frac{\sin \theta }{r}\right)$ (24)

$\frac{\partial P}{\partial y}=\frac{\partial P}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial P}{\partial \theta }\frac{\partial \theta }{\partial y}=\frac{\partial P}{\partial r}\sin \theta +\frac{\partial P}{\partial \theta }\frac{\cos \theta }{r}$ (25)

二、将(22)和(23)代入(24)和(25)，得到

$\begin{array}{c}\frac{\partial P}{\partial x}=-\frac{\left(r^2-1\right)\sin \left(\psi -\theta \right)\left[2r^3\cos \left(\psi -\theta \right)+r^4+10r\cos \left(\psi -\theta \right)-10r^2-3\right]}{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\\ +\frac{-6r^5{\cos }^2\left(\psi -\theta \right)\cos \theta +r^6\cos \theta \cos \left(\psi -\theta \right)-8r^3{\cos }^2\left(\psi -\theta \right)\cos \theta }{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\\ +\frac{17r^4\cos \left(\psi -\theta \right)\cos \theta }{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\frac{-10r{\cos }^2\left(\psi -\theta \right)\cos \theta +27r^2\cos \left(\psi -\theta \right)\cos \theta }{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\\ -\frac{\left(r^2-1\right)\sin \theta \sin \left(\psi -\theta \right)\left[2r^3\cos \left(\psi -\theta \right)+r^4+10r\cos \left(\psi -\theta \right)-10r^2-3\right]}{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\\ +\frac{-20r^3\cos \theta +3\cos \left(\psi -\theta \right)\cos \theta -4r\cos \theta }{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\end{array}$ (26)

$\begin{array}{c}\frac{\partial P}{\partial y}=\frac{\partial P}{\partial r}\sin \theta +\frac{\partial P}{\partial \theta }\frac{\cos \theta }{r}\\ =\frac{-6r^5{\cos }^2\left(\psi -\theta \right)\sin \theta +r^6\cos \left(\psi -\theta \right)\sin \theta -8r^3{\cos }^2\left(\psi -\theta \right)\sin \theta }{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\\ +\frac{17r^4\cos \left(\psi -\theta \right)\sin \theta }{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}+\frac{-10r{\cos }^2\left(\psi -\theta \right)\sin \theta +27r^2\cos \left(\psi -\theta \right)\sin \theta }{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\\ +\frac{\left(r^2-1\right)\cos \theta \sin \left(\psi -\theta \right)\left[2r^3\cos \left(\psi -\theta \right)+r^4+10r\cos \left(\psi -\theta \right)-10r^2-3\right]}{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\\ +\frac{-20r^3\sin \theta +3\cos \left(\psi -\theta \right)\sin \theta -4r\sin \theta }{2{\left[1-2r\cos \left(\psi -\theta \right)+r^2\right]}^3}\end{array}$ (27)

其中 [8]

${\left|z-\zeta \right|}^2=1-2r\cos \left(\theta -\psi \right)+r^2$ (28)

$\cos \left(\theta -\psi \right)=\frac{1+r^2-{\left|z-\zeta \right|}^2}{2r}$ (29)

${\cos }^2\left(\theta -\psi \right)=\frac{{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}{4r^2}$ (30)

$\sin \left(\theta -\psi \right)=\sqrt{1-{\cos }^2\left(\theta -\psi \right)}=\sqrt{1-\frac{{\left(1-{\left|z-\zeta \right|}^2+r^2\right)}^2}{4r^2}}$ (31)

$\nabla P\left(z,\zeta \right)=\left(\frac{\partial P}{\partial x},\frac{\partial P}{\partial y}\right)$ (32)

我们设 $l=\left(1,0\right)$ ，则

$\begin{array}{c}\langle \nabla P\left(z,\zeta \right),l\rangle =\frac{-3r^3{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2\cos \theta +r^5\cos \theta \left(1+r^2-{\left|z-\zeta \right|}^2\right)}{4{\left|z-\zeta \right|}^6}\\ -\frac{4r\cos \theta {\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}{4{\left|z-\zeta \right|}^6}+\frac{27r\cos \theta \left(1+r^2-{\left|z-\zeta \right|}^2\right)}{4{\left|z-\zeta \right|}^6}\\ +\frac{17r^3\cos \theta \left(1+r^2-{\left|z-\zeta \right|}^2\right)-5\left[\frac{{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}{r}\right]\cos \theta }{4{\left|z-\zeta \right|}^6}\\ +{\frac{-40r^3\cos \theta +3\left(\frac{1+r^2-{\left|z-\zeta \right|}^2}{r}\right)\cos \theta -4r\cos \theta }{4{\left|z-\zeta \right|}^6}}_{{}_{}}\end{array}$

$\begin{array}{l}+\frac{\frac{r^2-1}{r}\sin \theta \sqrt{4r^2-{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}\left[5\left(1+r^2-{\left|z-\zeta \right|}^2\right)\right]}{4{\left|z-\zeta \right|}^6}\\ +\frac{\frac{r^2-1}{r}\sin \theta \sqrt{4r^2-{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}\left[r^2\left(1+r^2-{\left|z-\zeta \right|}^2\right)\right]}{4{\left|z-\zeta \right|}^6}\\ +\frac{\frac{r^2-1}{r}\sin \theta \sqrt{4r^2-{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}\left(r^4-10r^2-3\right)}{4{\left|z-\zeta \right|}^6}\end{array}$ (33)

由(33)，我们得到最后的定理

定理2.3设 $\zeta ={\text{e}}^{i\psi }$ ， $z=r{\text{e}}^{i\theta }$ ，当 $l=\left(1,0\right)$ 时，有

$\begin{array}{c}C\left(z,l\right)={\displaystyle {\int }_{\partial D}\frac{17r^3\cos \theta \left(1+r^2-{\left|z-\zeta \right|}^2\right)-5\left[\frac{{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}{r}\right]\cos \theta }{4{\left|z-\zeta \right|}^6}}\\ \frac{-3r^3{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2\cos \theta +r^5\cos \theta \left(1+r^2-{\left|z-\zeta \right|}^2\right)}{4{\left|z-\zeta \right|}^6}\\ -\frac{4r\cos \theta {\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}{4{\left|z-\zeta \right|}^6}+\frac{27r\cos \theta \left(1+r^2-{\left|z-\zeta \right|}^2\right)}{4{\left|z-\zeta \right|}^6}\\ +\frac{-40r^3\cos \theta +3\left(\frac{1+r^2-{\left|z-\zeta \right|}^2}{r}\right)\cos \theta -4r\cos \theta }{4{\left|z-\zeta \right|}^6}\end{array}$

$\begin{array}{l}+\frac{\frac{r^2-1}{r}\sin \theta \sqrt{4r^2-{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}\left[5\left(1+r^2-{\left|z-\zeta \right|}^2\right)\right]}{4{\left|z-\zeta \right|}^6}\\ +\frac{\frac{r^2-1}{r}\sin \theta \sqrt{4r^2-{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}\left[r^2\left(1+r^2-{\left|z-\zeta \right|}^2\right)\right]}{4{\left|z-\zeta \right|}^6}\\ +\frac{\frac{r^2-1}{r}\sin \theta \sqrt{4r^2-{\left(1+r^2-{\left|z-\zeta \right|}^2\right)}^2}\left(r^4-10r^2-3\right)}{4{\left|z-\zeta \right|}^6}\end{array}$ (34)

3. 结论

本文通过分析双解析函数和双调和函数的关系，进一步计算其Poisson核得到了双调和函数梯度范数的一个估计。由此，该估计丰富了双调和函数的应用范围，也为今后探究双调和函数的Khavinson猜想提供了一个积分表达式，给验证双调和函数的Khavinson猜想打下基础。